Let A be the region enclosed by the parabola $$y^2 = 2x$$ and the line $$x = 24$$. Then the maximum area of the rectangle inscribed in the region A is _____

The region A is enclosed by $$y^2 = 2x$$ and $$x = 24$$.

Consider a rectangle inscribed in this region with vertices at $$(x_0, y_0)$$, $$(24, y_0)$$, $$(24, -y_0)$$ and $$(x_0, -y_0)$$. Since the left side lies on the parabola, $$y_0^2 = 2x_0$$.

Hence the width is $$24 - x_0 = 24 - \frac{y_0^2}{2}$$ and the height is $$2y_0$$, so the area is

$$A = 2y_0\left(24 - \frac{y_0^2}{2}\right) = 48y_0 - y_0^3.$$

To maximize the area, differentiate with respect to $$y_0$$:

$$\frac{dA}{dy_0} = 48 - 3y_0^2 = 0,$$

which yields $$y_0^2 = 16\implies y_0 = 4.$$

A check of the second derivative, $$\frac{d^2A}{dy_0^2} = -6y_0$$, shows it is negative at $$y_0 = 4$$, confirming a maximum.

Substituting back into the area expression gives $$48(4) - 64 = 192 - 64 = 128.$$

The answer is $$\boxed{128}$$.